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JIPMER 2004
Exam
JIPMER
Year
2004
Subject
Physics
Chapter
Work, Power and Energy

JIPMER 2004 · Physics · Work, Power and Energy

A particle moves from position r _1=3 i +2 j -6 k to position r...

A particle moves from position r→1=3i^+2j^−6k^\overrightarrow{\mathbf{r}}_1=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} to position r→2=14i^+13j^+9k^\overrightarrow{\mathbf{r}}_2=14 \hat{\mathbf{i}}+13 \hat{\mathbf{j}}+9 \hat{\mathbf{k}} under action of force F=4i^+j^+3k^N\mathbf{F}=4 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \mathrm{N}. The work done will be:

Options

  1. A100 J
  2. B50 J
  3. C200 J
  4. D75 J

Correct answer

A. 100 J

Step-by-step solution

Using the formula for work done W=F→⋅r→=F→⋅(r2→−r1→)W=\overrightarrow{\mathbf{F}} \cdot \overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{F}} \cdot\left(\overrightarrow{\mathbf{r}_2}-\overrightarrow{\mathbf{r}_1}\right)W=(4i^+j^+3k^)⋅(14i^+13j^+9k^)−3i^−2j^+6k^)\begin{array}{r}W=(4 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(14 \hat{\mathbf{i}}+13 \hat{\mathbf{j}}+9 \hat{\mathbf{k}}) -3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})\end{array}W=(4i^+j^+3k^)⋅(11i^+11j^+15k^)W=44+11+45=100 joule \begin{aligned}& W=(4 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(11 \hat{\mathbf{i}}+11 \hat{\mathbf{j}}+15 \hat{\mathbf{k}}) \\ & W=44+11+45=100 \text{ joule }\end{aligned}$$
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